So far my difficulty is trying to find a $\delta$ that will allow for this function to be less than $\epsilon$. I keep getting that $|\sqrt{9-x}+3-6|=|\sqrt{9-x}-3| < \epsilon$, but from here I do not know how to deal with finding $\delta$. Thanks in advance and any help is appreciated.

If you have results about continuous functions you're allowed to use, then by far the easiest is to note that $f(x)=\sqrt{9-x}+3$ is continuous in a neighborhood of $0$ (namely $(-\infty,9)$) and therefore $\lim_{x\to 0}f(x) = f(0)$.
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Henning MakholmOct 4 '12 at 16:15

Sorry I don't know that result, or anything about continuous functions. All I know is the definition of a $\epsilon-\delta$ definition of a limit.
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tk2Oct 4 '12 at 16:17